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Find the derivative of \[\cos \left( \log x+{{e}^{x}} \right)\].

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Hint:For the above question we have been given a composite function to find its derivative and we know that to find the derivative of a composite function we will have to use the chain rule of derivative which states that the derivative of a composite function,
\[F\left( g\left( x \right) \right)\] is \[F'\left( g\left( x \right) \right)\times g'(x)\]

Complete step-by-step answer:
We have been asked to find the derivative of the function, \[\cos \left( \log x+{{e}^{x}} \right)\].
Since it is a composite function, so we will use the chain rule of derivative to find its derivatives and the chain rule states that the derivative of a function \[F\left( g\left( x \right) \right)\] is equal to \[F'\left( g\left( x \right) \right)\times g'(x)\].
So the differentiation of \[\cos \left( \log x+{{e}^{x}} \right)\] is shown as follows by applying the chain rule:
\[\dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log +{{e}^{x}} \right)\times \dfrac{d}{dx}\left( \log x+{{e}^{x}} \right)\]
Since we know that the derivative of cos x is –sin x.
\[\Rightarrow -\sin \left( \log x+{{e}^{x}} \right)\times \left[ \dfrac{d}{dx}\left( \log x \right)+\dfrac{d}{dx}\left( {{e}^{x}} \right) \right]\]
As we know the derivative of log x is \[\dfrac{1}{x}\] and the derivative of \[{{e}^{x}}\] is \[{{e}^{x}}\], we can write the above function as follows.
\[\Rightarrow \dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\]
Therefore, the required derivative of the given function is equal to \[-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\].
Note: Be careful while applying the chain rule to the composite function and also take care of the sign. Sometimes we take the derivative of cos x as sin x by mistake and we just forget the negative sign before sin x, so be careful at that time otherwise we will get incorrect answers.